Let us assume we have two parameters, $p_1$ and $p_2$. We also have two maximum likelihood estimators $\hat{p_1}$ and $\hat{p_2}$ and two confidence intervals for these parameters. Is there a way to build a confidence interval for $p_1p_2$?

3 Answers
3

You can use the Delta method to calculate the standard error of $\hat{p_{1}}\hat{p_{2}}$. The delta method states that an approximation of the variance of a function $g(t)$ is given by:
$$
\mathrm{Var}(g(t))\approx \sum_{i=1}^{k}g'_{i}(\theta)^{2}\mathrm{Var}(t_{i})+2\sum_{i>j}g'_{i}(\theta)g'_{j}(\theta)\mathrm{Cov}(t_{i},t_{j})
$$
The approximation of the expectation of $g(t)$ on the other hand is given by:
$$
\mathrm{\mathbf{E}}(g(t))\approx g(\theta)
$$
So the expectation is simply the function. Your function $g(t)$ is: $g(p_{1}, p_{2})=p_{1}p_{2}$. The expectation of $g(p_{1}, p_{2})=p_{1}p_{2}$ would simply be: $p_{1}p_{2}$. For the variance, we need the partial derivatives of $g(p_{1}, p_{2})$:
$$
\begin{align}
\frac{\partial}{\partial p_{1}}g(p_{1}p_{2}) & = p_{2} \\
\frac{\partial}{\partial p_{2}}g(p_{1}p_{2}) &= p_{1} \\
\end{align}
$$

Using the function for the variance above, we get:

$$
\mathrm{Var}(\hat{p_{1}}\hat{p_{2}})=\hat{p_{2}}^{2}\mathrm{Var}(\hat{p_{1}}) + \hat{p_{1}}^{2}\mathrm{Var}(\hat{p_{2}})+2\cdot \hat{p_{1}}\hat{p_{2}}\mathrm{Cov}(\hat{p_{1}},\hat{p_{2}})
$$
The standard error would then simply be the sqare root of the above expression. Once you've got the standard error, it is straight forward to calculate a 95% confidence interval for $\hat{p_{1}}\hat{p_{2}}$: $\hat{p_{1}}\hat{p_{2}}\pm 1.96\cdot \widehat{\mathrm{SE}}(\hat{p_{1}}\hat{p_{2}})$

To caluclate the standard error of $\hat{p_{1}}\hat{p_{2}}$, you need the variance of $\hat{p_{1}}$ and $\hat{p_{2}}$ which you usually can get by the variance-covariance matrix$\Sigma$ which would be a 2x2-matrix in your case because you have two estimates. The diagonal elements in the variance-covariance matrix are the variances of $\hat{p_{1}}$ and $\hat{p_{2}}$ while the off-diagonal elements are the covariance of $\hat{p_{1}}$ and $\hat{p_{2}}$ (the matrix is symmetric). As @gung mentions in the comments, the variance-covariance matrix can be extracted by most statistical softwares. Sometimes, estimation algorithms provide the Hessian matrix (I won't go into details about that here), and the variance-covariance matrix can be estimated by the inverse of the negative Hessian (but only if you maximized the log-likelihood!; see this post). Again, consult the documentation of your statistical software and/or the web on how to extract the Hessian and on how to calculate the inverse of a matrix.

Alternatively, you can get the variances of $\hat{p_{1}}$ and $\hat{p_{2}}$ from the confidence intervals in the following way (this is valid for a 95%-CI): $\mathrm{SE}(\hat{p_{1}})=(\text{upper limit} - \text{lower limit})/3.92$. For an $100(1-\alpha)\%$-CI, the estimated standard error is: $\mathrm{SE}(\hat{p_{1}})=(\text{upper limit} - \text{lower limit})/(2\cdot z_{1-\alpha/2})$, where $z_{1-\alpha/2}$ is the $(1-\alpha/2)$ quantile of the standard normal distribution (for $\alpha=0.05$, $z_{0.975}\approx 1.96$). Then, $\mathrm{Var}(\hat{p_{1}}) = \mathrm{SE}(\hat{p_{1}})^{2}$. The same is true for the variance of $\hat{p_{2}}$. We need to covariance of $\hat{p_{1}}$ and $\hat{p_{2}}$ too (see paragraph above). If $\hat{p_{1}}$ and $\hat{p_{2}}$ are independent, the covariance is zero and we can drop the term.

$\begingroup$@gung On a point of pedantry it might be worth pointing out (because I know it confuses some people) that it's really the variance-covariance matrix of $\hat \beta$ rather than $\beta$ (and in fact it isn't even really that, because the standard deviation has to be estimated from the sample, so it's really the estimated variance-covariance matrix..)$\endgroup$
– SilverfishDec 21 '15 at 16:32

If x and y are independently distributed, the variance of the product
is relatively straightforward:
V(x*y)= V(y)*E(x)^2 + V(x)*E(y)^2 + V(x)*V(y)
These results also generalize to cases
involving three or more variables (Goodman 1960).
Source: Regulating Pesticides (1980), appendix F

Coolserdash: The last component V(x)*V(y) is missing in your equation. Is the referenced book (Regulating Pesticides) wrong?

Also, both equations might not be perfect. "... we show that the distribution of the product of three independent normal variables is not normal." (source). I would expect some positive skew even in the product of two normally distributed variables.